Ta có:
1+\(\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{a}\)
Thay a=1
=>\(1+\dfrac{1}{b}=b+\dfrac{1}{c}=c+1\)
*Lấy \(1+\dfrac{1}{b}=c+1\Rightarrow\dfrac{1}{b}=c\Rightarrow b=\dfrac{1}{c}\)
=>\(1+\dfrac{1}{b}=\dfrac{2}{c}=c+1\)
*Lấy \(\dfrac{2}{c}=\dfrac{c+1}{1}\)
=> 2=c(c+1)
<=> 2=c2+c
=>c=-2
*Lấy \(1+\dfrac{1}{b}=\dfrac{2}{c}\)
Thay c=-2 và quy đồng
=>\(\dfrac{b+1}{b}=-1\)
=>b+1=-b
=> b+b=-1
=>2b=-1
=> b=-1/2
Vậy b=\(-\dfrac{1}{2};c=-2\)